SHCT Oral Exam

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Last updated 6:45 PM on 8/11/26
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67 Terms

1
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Please give us a short review of the problem.

We designed and dimensioned a server-room cooling system to maintain roughly 15°C, using ambient-air ventilation for free cooling and active AC cooling, during 4 representative days

We compared three refrigerants and three compressor bore sizes.

we considered overall energy demand, number of start/stop cycles, operating times, and temp and humidity in the room

2
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Briefly describe your solving approach.

We divided the system into components: room, ventilation, controller, AC cycle, compressor, and humidity. We developed governing equations and constraints for each, implemented them separately, integrated them, and simulated operation at 5-minute timesteps for four representative seasonal days.

3
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What were your main results?

Propane + 40 mm bore was the best combination, with the lowest electricity use, cost, and CO₂ emissions. Winter used only free cooling, spring/fall were ventilation dominated, and summer was AC dominated. AC accounted for about 79% of annual cooling electricity.

4
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How did you compare all designs?

We used nested loops over refrigerant, compressor bore, and season, then compared electricity use, cost, CO₂ emissions, cycling, runtime, temperature, and humidity.

5
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Why was the 40 mm compressor better than the 50 mm compressor?

The 50 mm provided more capacity, but caused shorter cycles, more frequent starts/stops, higher energy consumption, and larger temperature extremes.

6
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Why weren't the 30 mm compressors simulated?

At Troom​=15∘C and Tamb​=25∘C, they could not provide the required 5 kW peak cooling capacity.

7
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Why was minimum AC capacity based on 5 kW?

5 kW was the peak server heat load, so we required the AC to be capable of removing that load at the chosen reference condition.

8
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Why did you think sizing the AC from peak server load was reasonable?

HVAC systems are commonly sized based on peak expected cooling demand, so maximum server heat generation gave a logical first sizing criterion.

9
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Wouldn't checking the AC across the entire operating range be better?

Yes. A compressor capable of 5 kW at the reference condition may not provide 5 kW at hotter ambient conditions. Ideally, capacity should be checked across the full operating envelope.

10
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What happens if you don't eliminate compressors below 5 kW before simulation?

The 30 mm compressors would be simulated. They might use lower instantaneous power and have less cold overshoot, but would likely run longer and produce higher peak temperatures because they cannot meet peak loads

11
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Why does compressor bore affect cooling capacity?

A larger bore increases displacement, so the compressor moves more refrigerant per cycle, increasing refrigerant mass flow and cooling capacity.

12
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Why doesn't maximum COP necessarily mean minimum annual electricity use?

Annual energy also depends on capacity, runtime, cycling, fan energy, and how often each operating condition occurs.

13
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How did you calculate room temperature?

From an explicit euler step (next temp is current temp + temp rate of change times*delta T), using the energy balance of the server, vent, and AC

14
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Why did you assume constant air density?

The project description treated room volume and air mass as constant, which simplified the energy balance. However, because humidity changed, constant density was an imperfect assumption.

15
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Why did you assume constant cp​ for air?

It simplified the transient energy balance and avoided repeated property calculations.

16
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Why did you think constant cp​ was reasonably justified?

Air cp​ changes only slightly over the simulated temperature range, so its effect should be small.

17
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Wouldn't variable air density and cp​ be better?

Yes. They could be calculated from the current temperature and humidity using CoolProp, making the model more thermodynamically consistent.

18
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How do your results change if air density and cp​ are variable?

Heating/cooling rates, temperature peaks, switching times, electricity use, and cycling would change slightly, but the preferred design would probably remain unchanged.

19
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How did you calculate ventilation cooling?

from the relationship between mass flow rate, heat capacity, and ambient and room temp, with airflow limited by duct size and maximum velocity

20
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Why did you choose a 600 mm ventilation duct?

It gave a maximum flow of about 1.414 m³/s and 28 ACH, within the selected 20–40 ACH range while respecting the 5 m/s duct-velocity limit.

21
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Wouldn't optimizing the ventilation size be better?

es. Airflow or duct diameter could be optimized to minimize

Efan​+Ecompressor​

while maintaining temperature limits.

22
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Why wouldn't you simply maximize ventilation airflow?

More airflow increases free cooling but also increases fan electricity and could overcool the room. The optimum balances compressor savings against fan energy.

23
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Why is free cooling more efficient than AC?

It primarily requires fan electricity instead of compressor electricity, so when ambient air is sufficiently cold it uses much less energy.

24
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Why does increasing ventilation airflow increase cooling capacity?

increasing mass flow increases sensible cooling as long as ambient air is colder than the room

25
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Why does fan power increase strongly with fan speed?

power is proportional to the cube of the rotational speed so doubling a fan’s speed requires 8 times more power to operate

26
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How did the controller work in code?

We used if/elif logic based on room and ambient temperatures. Free cooling had priority; if ventilation could not provide sufficient cooling, AC was allowed to operate.

27
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Why did the controller primarily use room temperature?

Room temperature was treated as the primary indicator of cooling demand, with ventilation and AC switching at fixed temperature thresholds.

28
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Wouldn't controlling based on cooling demand be better than fixed temperature thresholds?

Yes. We could calculate:

Q˙​remaining​=Q˙​server​−Q˙​vent​

and use AC when ventilation cannot meet the remaining load. This considers where the temperature is heading, not just its current value.

29
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Why is hysteresis justified in an on/off controller?

Separate ON and OFF temperatures prevent rapid switching when room temperature fluctuates around the setpoint.

30
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What happens if the hysteresis band is too narrow?

The compressor cycles frequently, increasing cycling losses and wear.

31
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What happens if the hysteresis band is too wide?

Cycling decreases, but room-temperature fluctuations increase.

32
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How did you model compressor cycling?

We stored ON and OFF timers and updated them each timestep to enforce minimum run and standstill times.

33
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Why does compressor on/off control decrease average COP?

Start-up and shutdown introduce transient losses, and frequent cycling means more operation away from efficient steady-state conditions.

34
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Wouldn't variable-speed compression be better than on/off control?

Potentially. It could match capacity more closely to instantaneous load, reducing cycling and temperature oscillations and potentially improving part-load efficiency.

35
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How did you calculate COP and capacity efficiently?

We generated COP and capacity maps over room and ambient temperatures for each refrigerant/bore pair. During the simulation, we interpolated from these maps rather than recalculating the entire cycle every timestep.

36
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Why did you use analytical rather than numerical COP optimization?

Numerical optimization was much slower and sometimes failed to converge.

37
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Why did you think analytical COP optimization was reasonably justified?

We compared the approaches across representative room and ambient temperatures and generally obtained very similar COPs.

38
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Wouldn't numerical COP optimization be better?

Physically, yes. It could optimize more variables and handle interacting constraints. The disadvantage is increased computational cost and convergence problems.

39
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How do your results change with numerical COP optimization?

COP could improve at some operating points because more realistic constraints and additional decision variables could be included.

40
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Why does increasing pressure ratio decrease COP?

Higher pressure ratio requires more compressor work, so COP generally decreases.

41
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Why does lower evaporating temperature decrease COP?

It increases the temperature lift and pressure ratio, increasing compressor work.

42
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Why does higher condensing temperature decrease COP?

It increases condenser pressure and therefore compressor work.

43
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Why is compressor isentropic efficiency important?

Real compression requires more work than ideal isentropic compression. Lower isentropic efficiency therefore gives higher electricity consumption and lower COP.

44
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Why did you optimize evaporating temperature while essentially fixing condensing temperature?

The fixed condenser approach meant Tcond​ was essentially determined by ambient temperature. We therefore increased Tevap​ as much as constraints allowed because this generally increases COP.

45
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Wouldn't optimizing Tevap​ and Tcond​ together be better?

Yes. Changing evaporating temp changes compressor mass flow, pressure ration, operating limits, cooling cap

 

Increasing T evap also means compressor pumps less refrigerant, cooling capacity changes, and condenser operating conditions change, so may not actually remove enough heat from the room

46
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How do your results change if Tevap​ and Tcond​ are optimized together?

COP could increase at some operating points, compressor electricity could decrease, and cooling capacity could change, especially when the minimum pressure-ratio constraint is active.

47
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Why can't Tevap​=Troom​ or Tcond​=Tambient​?

Heat transfer requires a finite temperature difference. Refrigerant must evaporate below room temperature and condense above ambient temperature.

48
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Why does reducing heat-exchanger approach temperature improve COP?

It allows warmer evaporation and/or cooler condensation, reducing temperature lift and compressor work.

49
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Why wouldn't you make the approach temperature as small as possible?

A smaller approach requires a larger or more effective heat exchanger, so improved COP comes at the cost of greater heat-exchanger size/cost.

50
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Why did you assume fixed superheat and subcooling?

We followed the class-cycle model using fixed ΔTSH​ and ΔTSC​, reducing the number of optimization variables.

51
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Wouldn't variable superheat and subcooling be better?

Yes. Their optimum values depend on refrigerant and operating conditions and affect mass flow, compressor work, cooling capacity, and COP.

52
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Why is superheating necessary?

It ensures vapor rather than liquid refrigerant enters the compressor, protecting the compressor from liquid damage.

53
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Why is subcooling useful?

It lowers liquid refrigerant enthalpy before expansion, so less flash vapor forms (evaporation due to the sudden pressure drop) which means more evaporation can occur in the evaporator for better cooling

54
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Why did you approximate evaporator outlet temperature using evaporating temperature?

It was a simplifying assumption used in class exercises and avoided requiring a detailed evaporator model.

55
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Why was that approximation considered reasonable?

Evaporating temperature provides a reasonable first approximation of the cold coil temperature.

56
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Wouldn't explicitly modeling the evaporator outlet be better?

Yes. It would improve predictions of air outlet conditions, condensation, latent cooling, and humidity.

57
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How did you model humidity in code?

We tracked humidity ratio as a state variable. Ventilation moved it toward outdoor humidity, AC reduced it when condensation occurred, and moist fluid cp converted humidity ratio and temperature back to RH.

58
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Why did you calculate humidity separately from the energy balance?

Temperature control was the primary objective and humidity was treated mainly as a secondary reporting variable.

59
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Wouldn't coupling humidity to the energy balance be better?

Yes. Condensation removes latent heat, so it should contribute to AC cooling demand.

60
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How do your results change if humidity is included in the energy balance?

Cooling demand, compressor runtime, and electricity increase because the AC must remove both sensible and latent heat. Temperature undershoot would likely decrease.

61
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Which designs would be most affected?

Large compressors under humid conditions, because their greater capacity can cool below the dew point more quickly and cause more condensation.

62
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Why does ignoring humidity underestimate cooling demand?

Condensation requires latent heat removal. A sensible-only energy balance ignores this additional evaporator load.

63
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Why can relative humidity increase even if no water is added?

Cooling lowers the saturation vapor pressure, so the same absolute moisture content corresponds to a higher relative humidity.

64
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Why did you convert 30-minute weather data to constant 5-minute values?

The simulation required 5-minute timesteps but measurements were only available every 30 minutes. Holding values constant avoided inventing intermediate measurements.

65
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Why was this considered reasonable?

It preserved the actual measured values rather than introducing artificial interpolated values.

66
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Wouldn't interpolating the weather data be better?

Potentially. Linear interpolation would produce smoother ambient changes and more realistic controller operation.

67
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How did the main simulation work?

We looped through 5-minute timesteps, updating the controller, ventilation, AC cycle, room temperature, and humidity at each step.